So 5 choose 3 = **10 possible combinations**.

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## How many combinations of 3 numbers are there in 5 numbers?

We conclude that there are 35−3×25+3=**150 combinations**. If there were no restrictions, the number of five digit codes that could be formed with the digits 1, 2, 3 would be 35 since there are three ways to choose each of the five digits.

## How many permutations are there of 5 things taken 3 at a time?

Using the permutation formula:

[X]

The problem involves 5 things (A, B, C, D, E) taken 3 at a time. There are **60 different permutations** for the license plate.

## How many combinations of two items can be selected from a group of 5?

…n objects is called a permutation of n things taken r at a… (For k = n, _{n}P_{k} = n! Thus, for 5 objects there are 5! = **120 arrangements**.)

## How many combinations are possible with 3 items?

3*3*3=**27 unique possibilities**. This number is small enough to enumerate the possibilities to help your understanding (like the other tutors did), but the digits^base expression (with “^” meaning exponentiation) is important.

## How many 3 digit numbers can be formed using 12345?

As repetition is allowed, So the number of digits available for Y and Z will also be 5 (each). Thus, The total number of 3-digit numbers that can be formed = 5×5×5 = **125**.

## How many combinations can you make with 3 numbers without repeating?

If what you want are all possible three digit numbers with no repetition of the digits then you have 10 choices for the first digit, you have 9 choices for the 2nd digit, and you have 8 choices for the 3rd digit giving you 10x9x8 = **720** in all.

## How many ways can 3 objects be arranged?

A permutation is an arrangement of a number of objects in a definite order. There is a choice of 3 letters for the first place, then there is a choice of 2 letters for the second place and there is only 1 choice for the third place. Thus the three operations can be performed in 3 2 x 11=**6 ways**.

## How do you find possible combinations?

Remember that combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the **formula nCr = n! / r!** *** (n – r)!**, where n represents the number of items, and r represents the number of items being chosen at a time.

## How many combinations can you have with 3 teams?

Here is the “slick” way to solve it: there are 3 outcomes for each game (either the odd team wins, they tie, or the even team wins), and there are 3 separate games, so since each game is independent of the other, there are **33=27 possible outcomes**.

## How many permutations of 3 are there?

There are, you see, 3 x 2 x 1 = **6 possible ways** of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times. So we just divide by 6.

## How many combinations of 5 items are there?

Note that your choice of 5 objects can take any order whatsoever, because your choice each time can be any of the remaining objects. So we say that there are 5 factorial = 5! = 5x4x3x2x1 = **120 ways** to arrange five objects.

## How many ways can you choose 3 people in a group of 8?

When we computed **336 possible combinations**, we’ve counted each of them 6 times. 336/6 = 56, and that’s the answer. In general this number is called “8 choose 3” and would be denoted C(8,3). It’s the number of 3-element subsets of an 8-element set.

## How many combinations of 5 options are there?

For example, 5! = 5×4×3×2×1 = **120**. The number of ways to order a set of items is a factorial.

## How many combinations of 5 numbers are there?

Assuming no five-digit number can begin with zero, there are 9 possible choices for the first digit. Then there are 10 possible choices for each of the remaining four digits. Therefore, you have 9 x 10 x 10 x 10 x 10 combinations, or 9 x 10^4, which is **90,000 different combinations**.

## How many 3 digit numbers can be formed from the digits 12345 which are divisible by 3?

When we talk of divisibility by 3, on an average 1 out of every 3 numbers is divisible by 3, therefore, we will have 180 / 3 = **60 numbers** that would be divisible by 3. Thus, 60 is the required answer.

## How many 3 digit numbers can be formed using the first five counting numbers?

The answer is **125**. Assume one available position for each digit that can be filled by either of 1–5. Thus, if you are allowed to have numbers with repeated digits- such as 222,343, … – you will have 5*5*5=125 different numbers, otherwise 5*4*3=60.

## How many 3 letter combinations can you make with the alphabet?

There are **2,600 different combinations** of 3 letters in the alphabet. If you allow repeated characters, there are 3,276 different combinations. A lot. ( 17576 ) that’s according to google.

## How many 3 digit numbers can be formed using 2 3 4 and 5 None of the digit repeated?

Answer: There are **24 three digits numbers**.

## How many combinations are there with 3 letters and 3 numbers?

Now to combine this numbers; for each combination of the 3 letters, we have all the 1000 digit combinations available, which means we can multiply the total number of letter combinations (17576) with the total number of digit combinations (1000), which is **17,576,000** (seventeen million), this is quite a lot but not …

## How many 3-digit numbers can be formed if repetition is allowed?

Originally Answered: how many 3-digit numbers can be formed if repetition is allowed? 9*10*10=**900**.

## How many ways can you choose 3 things from a set of 7 things?

Now consider choosing 3 things. There are 7 ways of choosing the first, 6 ways of choosing the second and 5 ways of choosing the third. 7*6*5=**210**.

## How many ways can 6 items be arranged?

So the six letters can be a combination of 6×5×4×3×2×1 letters or **720 arrangements**.

## How many ways can 5 letters be arranged?

This is simply 5! =**120 different ways**.

## How many combinations of 6 items are there?

For any combination of items, each item is either included or not included in the combo. That means each item has 2 possibilities for every combination. For 6 items, that would make the number of combinations = 2^6 = **64**.

## How many combinations of 4 items are there?

I.e. there are 4 objects, so the total number of possible combinations that they can be arranged in is 4! = 4 x 3 x 2 x 1 = **24**.

## How do you calculate 6C3?

Mathematically nCr=n! r! ×(n−r)! Hence **6C3=6**!

## What is 5c2 worth?

5 CHOOSE 2 = **10 possible combinations**. 10 is the total number of all possible combinations for choosing 2 elements at a time from 5 distinct elements without considering the order of elements in statistics & probability surveys or experiments.

## How many combinations of 5 football games are there?

If only considering wins/losses, and ignoring the possibility of cancellation or forfeiture or other similar non-results, each match can be a win, loss, or tie from the perspective of each team. So, 3 outcomes per match, and 5 matches means there are 3^5 combinations, or **243**.

## How many combinations of n items are there?

The number of combinations of n distinct objects, taken r at a time is: **C _{r} = n! / r!** (n – r)! Thus, 27,405 different groupings of 4 players are possible.

## How do you calculate 4P2?

∙nPr=n! (n−r)! **4P2=4!** **(4−2)**!

## How many combinations of 3 games are there?

Similarly for 3 matches it would be, 3*3*3 = **27 possibilities**.

## How many combinations of 8 games are there?

Example: 8 games with 2 eventualities (victory, defeat) represent a total of 28=256 2 8 = **256 combinations**.

## How many ways are there to choose a committee of 3 people from a group of five people?

options for voting for exactly three candidates, 10 = 5-choose-2 = 5*4 / 2! for voting for two candidates, and 5 = 5-choose-1 options for voting for only one candidate. The grand total is **25 options** for voting for at least one and at most 3 out of 5 candidates.

## How many ways are there to choose a committee of 3 people from a group of 5 people group of answer choices?

Total # of ways: 5C3*2^3=**80**. Answer: D.

## How many combinations of 3 items can be chosen from a set of 4 items?

For how many combinations, you have it. Permutations is **24**.

## How many combinations of 5 items are there without repeating?

5 possibilities for 1st position x 4 for 2nd x 3 for 3rd x 2 for 4th x 1 for 5th = 120 different ways to arrange a 5-digit grouping without repeating digits. To remove duplicate 5-digit combinations: divide 6,375,600 by 120 = **56,130 five-**digit combinations possible without repetition of digits.

## How many combinations are there of 5 numbers without repeating?

So, The total number of 5 digit numbers have no digits repeated is 9x9x8x7x6 which is equal to **27216**. You can use whatever digits you wish except 0 for X1.

## How many 5 digit alphanumeric combinations are there?

Finally we can multiply them and get your answer which is 9*10*10*10*10 which results in **90000**.

## How many combinations are there with 5 numbers and letters?

If you can repeat letters there are **26** choices for each letter and 10 choices for each digit. In such a case there are 265×105 ways. If you can’t repeat letters then we have to arrange 5 letters from 26 possible letters and arrange 5 numbers from 10 numbers (the two choices are independent).